feedback systems - eecsweb.eecs.umich.edu/~jfr/embeddedctrls/files/lecture7.pdf · feedback systems...

Feedback Systems

Many embedded system applications involve the concept offeedback

Sometimes feedback is designed into systems:

CPU

Operator Input

Actuator PhysicalSystem

positionvelocitytemperaturecurrent...

Sensor

Other systems have naturally occuring feedback, dictated bythe physical principles that govern their operation

EECS461, Lecture 7, updated September 24, 2008 1

Feedback Systems

Some examples we will see:- op-amp

- motor equations: mechanical

- motor equations: electrical

- DC motor: back EMF

- current controlled amplifier

- velocity feedback control

How many examples of feedback can you think of?


Issues with Feedback

A feedback loop in a system raises many issues- requires a sensor!

- changes gain

- reduces effects of parameter uncertainty

- may alter stability

- changes both steady state as well as dynamic response

- introduces phase lag

- sensitive to computation/communication delay

Detailed analysis (and design) of feedback systems is beyondthe scope of our course, but we will need to understand these

basic issues...


Feedback and Gain

Using high gain in a feedback system can make output trackinput:

u y

-

GK

e

feedback response:

y =KG

1 + KGu

error response:e =

1

1 + KGu

high gain: as K , y u and e 0- open loop gain: |KG| 1- closed loop gain: |KG/(1 + KG)| 1 we can make the output track the input even if we don;t

know the exact value of the open loop gain!

CAVEAT: only useful if system is stable!- for all but very simple systems, use of excessively high gain

will tend to destabilize the system!

a simple example where dynamics are usually ignored: op amp


Operational Amplifier (Op Amp)

An op amp [2] is used in many electronics found in embeddedsystems. Hence it is of interest in its own right, as well as

being a simple example of a feedback system

-

+

v1

v2vo

v1, v2: input voltages

vo: output voltage

output voltage depends on difference of input voltages

vo = K(v2 v1) = K(v1 v2)

Typically K 105 106, but varies significantly due tomanufacturing tolerances

Ideal op amp- no current flows into input terminals

- output voltage unaffected by load

In reality- op amp is a low pass filter with very high bandwidth

- draws a little current

- is slightly affected by load

we shall assume an ideal op amp


Inverting Amplifier, I

Q: How to use the op amp as an amplifier given that gain isuncertain?

A: Feedback!

-

+vi

vo0

v

R1I1

R2I2

currents:I1 =

vi v

R1, I2 =

v voR2

feedback equations:- from previous page, vo depends on v

: vo = Kv- v depends on vo: v

= vo + R2I2

-K

-1

R2I2 vo

-

v


Inverting Amplifier, II

Feedback diagram:

-K

-1

R2I2 vo

-

v

Apply rule for transfer function of feedback system:

vo =

K

1 + K

R2I2

If K >> 1, then the feedback equations imply that

vo R2I2

It further follows that v = vo + R2I2 0. By assumptionthat the op amp draws no current, I1 = I2, and thus

vo =

R2

R1

vi

Feedback allows us to use an op amp to construct an amplifierwithout knowing the precise value of K!


More Complex Feedback Examples

to analyze op amp, we ignored dynamics and treated the opamp as a pure gain that was constant with frequency

in general, dynamics cannot be ignored- transient response

- stability

Two examples where feedback arises from the physics- motor dynamics: mechanical

- motor dynamics: electrical

we shall discuss these examples, but we will first consider asimple case: feedback around an integrator


Integrator

Equations of integrator

x = u

x(t) = x(0) +

Z t0

u()d

Examples:- u is velocity, x is position

- u is acceleration, v is velocity

- voltage and current through inductor: I = 1LR

V dt

- voltage and current through capacitor: V = 1CR

Idt

Integrator is an unstable system- the bounded input, u(t) = 1, yields the unbounded output

x(t) = x(0) + t

Transfer function of an integratorZ

1

s

integrator has infinite gain at DC, s = 0


Feedback Around an Integrator

Suppose there is feedback around integrator:x

.x

u -

K

differential equation of feedback system

x = Kx + Ku

Transfer function of feedback system:

xu

-

K

s

X(s) =

K/s

1 + K/s

U(s) =

K

s + K

U(s)

The system is stable if K > 0.

The response to the constant input u(t) = 1 yields

x(t) 1

x(t) 0

independently of the value of K


Uses of an Integrator

sometimes integrators arise from the physics other times they are constructed

- to perform analog simulation of physical system

- to add integral control to a system

Op-amp integrator

-

+

R1

C2

vo

vi

-

+

R3

R4

- Transfer function:

vo =R4

R3

1

R1C2svi

Can also implement integrator on a microprocessor- discrete simulations

- digital control


Motor Equations, Mechanical

equations of motion for shaft dynamics

J = TM B

=

1

J

TM

B

J

: shaft speed, B 0: friction coefficient, J > 0: shaftinertia, TM : motor torque

Feedback diagram

1

J

-

TM 1

s

B

.

Transfer function:

(s) =1

sJ

1 + BsJTM(s) =

1/B

sJ/B + 1TM(s)

Constant torque speed goes to a steady state value:

ss = TM/B

NOTE: with no friction (B = 0), system is unstable!- constant torque implies (t)


Motor Equations, Electrical

equations of armature winding (ignoring back emf)

LI = V RI

I =

1

L

V

R

L

I

I: current, R: resistance, J : inductance, V : applied voltage

Feedback diagram

1

L

-

V 1

s

R

.

Transfer function:

I(s) =1

sL

1 + RsLV (s) =

1/R

sL/R + 1V (s)

Constant voltage current goes to a steady state value:

Iss = V/R


First Order Systems

Shaft dynamics and circuit dynamics are each examples of afirst order systems; i.e., they each have one integrator

In general, a first order system may be written in the form

x = ax + bu

where x is the integrator state, u is the input, and a and b

are constants.

Feedback diagram:

ux

-

a

.x

b

Equivalently

ux

-

a/b

.x

b

Transfer function:

X(s) = H(s)U(s)

H(s) =

b

a

1

s/a + 1


Stability and Time Constant

Time response:

x(t) = eat

x(0) +

Z t0

ea(t)

bu()d

Response to a unit step, u(t) = 1, t 0:

x(t) =b

a

1 eat

The system is stable if a > 0

- stability implies that x(t) ba as t

Rate of convergence determined by time constant, = 1/a- at t = , step response achieves 63% of its final value

- at t = 2 , step response achieves 87% of its final value

- at t = 3 , step response achieves 95% of its final value

To easily compare rate of convergence, normalize so that b = a Normalized frequency response:

x = H(j)u, H(j) =

1

j + 1

NOTE: The time constant determines the rate at which the

response of the system must be sampled in order to adequately

represent it in digital form.


Bandwidth and Response Speed

Time constant, determines1

- bandwidth of frequency response:

102

101

100

101

102

70

60

50

40

30

20

10

0

mag

nitu

de, d

b

frequency response of first order filter, 1/(j+1)

= 0.1 = 1 = 10

102

101

100

101

102

100

80

60

40

20

0

, radians/second

phas

e, d

egre

es

- speed of response to unit step input, u(t) = 1:

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1step response of first order filter, 1/(j+1)

time, seconds

= 0.1 = 1 = 10

1Plots created with Matlab file first order.m.


Complete Motor Model

The motor has both electrical and mechanical components,interconnected by the back EMF feedback loop:

-

V-VBKM

KV

1

L

1s

R

.

1

J

-

TM 1

s

B

.

V

-

VB

Two integrators a second order system

Rules for combining transfer functions

(s) =

1

sL+R

KM

sJ+B

1 +

Kv

sL+R

KM

sJ+B

V (s)=

KM

(sJ + B)(sL + R) + KvKMV (s)


Second Order Systems

Question: How to analyze and describe properties of secondorder systems?

- stability

- steady state response

- transient response

Approach 1:- If the system can be decomposed into component first order

subsystems, then (perhaps) properties of the overall system

can be deduced from those of these subsystems.

- Example: DC motor

Approach 2: General analysis procedure.- Roots of characteristic equation

- Damping coefficient and natural frequency determine

response

- Example: Virtual spring/mass/damper systems

We will need to understand the relation between transientresponse and characteristic roots (natural frequency and damping)

in order to design force feedback algorithms in Lab 6!


Time Scale Separation

For a DC motor, the time constants for each first ordersubsystem may be very different:

- electrical subsystem: e = L/R = 0.001

- mechanical subsystem: m = J/B = 0.35

Mechanical subsystem is much slower than the electricalsubsystem

- Response of motor shaft is dominated by the mechanical

subsystem

- On the shaft speed time scale, current appears to be

instantaneous

- Since current and torque are related directly, TM = KMI,

torque also responds rapidly2

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.05

0.1

0.15

0.2

0.25

0.3

0.35response of DC motor to a unit step in voltage input

time, seconds

speed torque

2Matlab files motor linear.m and DC motor linear.mdl



Electrical dynamics can be ignored by setting L = 03

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.05

0.1

0.15

0.2

0.25

0.3


time, seconds

speed, L = 0 torque, L = 0 speed, L = 0.01 torque, L = 0.01

Detail:

0.95 0.96 0.97 0.98 0.99 1 1.01 1.02 1.03 1.04 1.050

0.02

0.04

0.06

0.08

0.1


time, seconds

L = 0 L = 0.01

Will need to model current when we implement torque control3Matlab files motor neglect circuit.m and DC motor linear.mdl



Systems with two integrators- DC motor

- system with input and output described by the differential

equation

y + by + ay = cu

-

u

-

c

y

a

.y

b

..y

The frequency response function can be written as

H(s) =c

s2 + bs + a

Example: DC Motor

H(s) =

KMJL

s2 +`

BL+JRJL

s +

BR+KMKV

JL


Characteristic Roots

Suppose the frequency response is given by

H(s) =c

s2 + bs + a

Define the characteristic equation:

s2+ bs + a = 0

Characteristic roots

s =b

b2 4a

2(1)

Possibilities:(i) b2 4a > 0 two distinct real roots

s-plane

xx

(ii) b2 4a = 0 one repeated real root

s-plane

x+

(iii) b2 4a < 0 two complex conjugate roots

s-planex

x


Characteristic Roots and Stability

Second order system is- stable if the characteristic roots lie in the Open Left Half

Plane (OLHP)

- unstable if the characteristic roots lie in the Closed Right

Half Plane (CRHP)

- (roots on the imaginary axis are sometimes called marginally

stable)

stable unstable

s-plane


Natural Frequency and Damping

Parameterize roots of s2 + bs + a = 0 by

s = n jnp

1 2 (2)

where natural frequency, n, and damping coefficient, , are

defined by (compare (2) with (1))

b = 2n, a = 2n

roots lie on circle of radius n at an angle

= arctan /p

1 2

with the imaginary axis:

jn

+

+

Roots are- real if 2 1- complex and stable if 0 < < 1

- imaginary if = 0


Frequency and Time Response

Natural frequency, n and damping ratio, determine4

- bandwidth and peak of frequency response:

101

100

101

60

40

20

0

20

40

mag

nitu

de, d

bfrequency response of second order filter, 1/((j/

n)2 + 2(j/

n)+1)

= 1 = 0.2 = 0

101

100

101

200

100

0

100

200

/n

phas

e, d

egre

es

- speed and overshoot of unit step response:

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

step response of second order filter, 1/((j/n)2 + 2(j/

n)+1)

nt

= 1 = 0.2 = 0

4Plots created with Matlab m-file second order.m.


General Systems

The characteristic equation of an n-th order system will haven roots; these roots are either real, or they occur in complex

conjugate pairs.

The characteristic polynomial can be factored as

NRYi=1

(s + pi)

NC/2Yi=1

(s2+ bis + ai)

Each pair of complex roots may be written as

si =bi2

qb2i 4ai

2= xi jyi

and have natural frequency and damping defined from

si = ini jniq

1 2i

Hence and n can be computed from the real and imaginaryparts as

ni =q

x2i + y2i , i = xi/ni

Note: It often happens that the response of a high order systemis well approximated by one complex pair of characteristic

roots.


Spring/Mass/Damper System

K

BM

F

y

Newtons laws:

My + By + Ky = F

y = B

My

K

My +

F

M

Second Order System

-

F

-

1M

y

K

.y

B

..y

Transfer Function:

Y (s) =1M

s2 + BM s +KM

F (s)


Motor Control Strategies

Can conceive of controlling four signals associated with themotor

- input voltage, V

- shaft position,

- shaft velocity,

- torque, TM (equivalently, current, I)

-

V-VBKM

KV

1

L

1s

R

.

1

J

-

TM 1

s

B

.

V

-

VB

-

TL

1s

Issues:- Input (V ) vs. output (, , I) variables

- Open loop vs. feedback control (i.e., do we use sensors?)

- Effect of load torque

- Control algorithm (P, I, ...)

Motor control results in higher order systems (more than twointegrators)

Higher order systems- Can still define characteristic polynomial and roots

- Stability dictates that characteristic roots must lie in OLHP

- Integral control may still be used to obtain zero error

(provided that stability is present)

- More complex control algorithms may be required to obtain

stability


Voltage Control

Apply desired V (either with a linear or a PWM amplifier) Suppose there is a constant load torque, TL. Then steady

state speed and torque depend on the load:

=KMV RTLKMKV + RB

TM =KM(V B + KV TL)

KMKV + RB

Position

-

V-VBKM

KV

1

L

1s

R

.

1

J

-

TM 1

s

B

.

V

-

VB

-

TL

1s

Issues:- V is an input variable, and usually not as important as TM ,

, or

- Suppose we want to command a desired speed (or torque),

independently of load or friction

* Problem: usually load torque (and often friction) are

unknown

- Suppose we want to command a desired position

* Problem: no control at all over position!


Position Control, I

Suppose we want to control position We can use a sensor (e.g., potentiometer) to produce a voltage

proportional to position, and compare that to a commanded

position (also in volts).

1sL+R

1sJ+B

V-VBKM

KV

I

-

TM 1s

potentiometerradiansvolts

-

* (volts) errorP-I

an integral controller cannot stabilize the system. Instead usea proportional-integral (P-I) controller: 10 + 1s

responses of speed, torque, and position due to a unit stepcommand to position5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

time, seconds

speed torque position

5Matlab files motor position FB.m and DC motor position.mdl


Position Control, II

P-I control: if feedback system is stable, then error approacheszero, and position tracks desired value

Can implement analog P-I control using op amp circuit Control can also be implemented digitally using a

microprocessor

An encoder can be used instead of a potentiometer to obtaindigital measurement

PWM can be used instead of linear amplifier


Velocity Control, I

Using an analog velocity measurement, from a tachometer,and an analog integral controller, allows us to track velocity

controller 1sL+R

1sJ+B

radians/second

V-VBKM

KV

I

-

TM

tachometervolts

-

* (volts) error

VB

V

TL

-

despite the presence of an unknown load torque6

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

time, seconds

load torque, TL = 0.5

speed torque

6Matlab files motor speed FB.m and DC motor speed.mdl


Velocity Control, II

microprocessor control- use encoder measurement to generate digital velocity

estimate

- compare measured speed with desired speed

- feed error signal into digital integral controller

- generate PWM signal proportional to error

Note: Performance depends on the controller gain7. Considerthe difference between 10/s and 100/s:

0 1 2 3 4 5 6 7 8 9 100.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

time, seconds

speed, K = 10 torque, K = 10 speed, K = 100 torque, K = 100

Usually,excessively high gain leads to oscillatory response orinstability!

7Matlab files motor speed FB.m and DC motor speed.mdl


Torque Control

Using a measurement of current and an analog integralcontroller, allows us to track torqe, which is directly

proportional to current: TM = KMI

controller 1

sL+R

1

sJ+B

V-VBKM

KV

I

-

TM

-

I* I error

VB

V

TL

-

despite the presence of an unknown load torque8

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5

4

time, seconds

load torque, TL = 0.5

speed torque

Question: How does our lab setup implement torque control?8Matlab files and motor current FB.m and DC motor current.mdl


PWM Amplifier, I

Copley 4122D DC brush servo amplifier with PWM inputs [1] Two feedback control modes:

- velocity control (requires a tachomoter)

- torque (current) control

We use torque control so that we can provide force feedbackthrough our haptic interface

+HV

GND

+

1

2

3

4

5

J1

J1

MOTOR

ENCODER

+6

16

J2

10

13

12

11

GND

ENABLE

POS ENABLE

NEG ENABLE

PWM

DIR

VELOCITY MODE

Note: JP1 on pins 2-3 ( default )

J2 TACH

+

5

4

+HV

GND

+

1

2

3

4

5

J1

J1

MOTOR

ENCODER

+6

16

J2

10

13

12

11

GND

ENABLE

POS ENABLE

NEG ENABLE

PWM

DIR

TORQUE MODE

Note: JP1 on pins 2-3 ( default )

Notes

1. All amplifier grounds are common (J1-3, J1-4, J2-2, J2-7, and J2-10 ) Amplifier grounds are isolated from case &heatplate..

2. Jumper JP1 default position is on pins 2-3 for ground active /Enable input ( J2-11 )For /Inhibit function at J2-11 ( +5V enables ), move JP1 to pins 1-2

3. For best noise immunity, use twisted shielded pair cable for tachometer inputs. Twist motor and power cables and shield to reduce radiated electrical noise from pwm outputs.


PWM Amplifier, II

one-wire mode: 50% duty cycle corresponds to zerorequested torque

analog integral controller with anti-windup H bridge PWM amplifier 25 kHz PWM output

NEG ENABLE

POS ENABLE

ENABLE *

-

+-

+RH7

RH6 CH8

RH10 CH11

100PF

TACH AMPGv = 1

DIR

PWM

RH1

CURRENT LIMITSECTION

Gv = +HV10

PWM

MOSFET"H"

BRIDGE

STAGE

GND

GND

MOTOR +

MOTOR -

+HV

-

+

CH17RH15

CH1647K

47K

PEAK

CONT

PEAK

RH13

RH12

CH14

Voltage gain = 1

TIME

1K

33NF

1K

33NF

CURRENTMONITOR

CURRENTREF

1K

OUTPUTCURRENT

SENSE

+HV

DC / DCCONVERTER

-15

+15

GND

+FAULT OUTPUT+5V

CONTROLLOGIC

STATUS&

MOTOR

AUXILIARYDC OUTPUTS

SERVOPREAMP

CW

TACH LEAD

INTEGRATOR

CURRENTERROR

AMP

LED

GREEN = NORMALRED = FAULT

+15V

-15V

BALANCE50K

CASE GROUNDNOT CONNECTEDTO CIRCUIT GROUND

CASE MAY BE GROUNDED

MOMENTARY SWITCH CLOSURE RESETS FAULTWIRE RESET TO GROUND FOR SELF-RESET

+5

+15

-15

POWER GROUND AND SIGNAL GROUNDS ARE COMMON

J2 SIGNAL CONNECTOR

J1 MOTOR & POWER CONNECTOR

-+

VOLTAGEGAIN

+/-6V for+/-Ipeak

+/-6V at+/-Ipeak

60.4 K

100 K

10 MEG

4.8 kHz FILTER

4.8 kHz FILTER

RESET

10k

10k

INTEGRATOR RESET SWITCHES TURN ON WHEN AMP IS DISABLED

FOR SHIELDING

16

6

GND7

4

5

9

8

1

3

15

14

13

11

12

1

2

3

4

5

VALUE DEPENDS ON MODELSEE "ARMATURE INDUCTANCE" TABLE

2TACH(-) / GND

+15V

-15V

CH9 (OPEN)

*JP1

10

-

+

123

* JP1: PINS 2-3 FOR /ENABLE AT J2-11PINS 1-2 FOR /INHIBIT AT J2-11

0.47U

4.7MEG

182K

NOTE:

DEFAULT VALUE OF CH11 IS 0 OHMSFOR TORQUE MODE OPERATION

FOR VELOCITY MODE ( BRUSH TACH )SEE APPLICATIONS SECTION

CURRENT

CURRENT

*

*

PWMTO

+/-10vANALOG

CONVERSION

-

+

RH5

100 K

PWM BOARD

+/-10VMAX

TACH

Rext

Rin = 47K

LM-629

PWM MAG

PWM SIGN

19

18


References

[1] Copley Controls. Models 4122D, 4212D DC brush servo

amplifiers with PWM inputs. www.copleycontrols.com.

[2] K. Ogata. Modern Control Engineering. Prentice-Hall, 3rd

edition, 1997.


feedback systems - eecsweb.eecs.umich.edu/~jfr/embeddedctrls/files/lecture7.pdf · feedback systems...

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